# Budget Constraint in Utility Maximisation Problem with Lagrange Multipliers

Lets say we have a utility function $$U: \mathbb{R}^{2} \to \mathbb{R}$$ given by $$U(x,y)$$ and a binding budget constraint $$p_{x} x + p_{y} y = m$$, where $$p_{x}, p_{y}$$ are prices of goods $$x,y$$ and $$m$$ is income. We can maximize this function subject to our constraint by simply maximizing this the following function in the unconstrained sense: $$\mathcal{L}(x,y, \lambda) = U(x,y) + \lambda(m -p_{x}x - p_{y}{x})$$ where $$\lambda$$ is usually called the Lagrange Multiplier.

My question: The reason we're able to add the second term on the RHS of the expression above is because our budget constraint binds, i.e. $$m - p_{x}x - p_{y}y = 0$$, so we're essentially adding zero. But, why is the conventional expression inside the parantheses after $$\lambda$$ given by $$m - p_{x}x - p_{y}y$$ and not $$p_{x}x + p_{y}y - m$$? Shouldn't both expressions give us a zero term? My guess is that this has something to do with the sign of $$\lambda$$ (which we want to be positive because it can be interpreted as the marginal utility of $$m$$), but I'm not sure how exactly this works, mathematically. Would appreciate a clarification explained mathematically.

Objective function of Lagrangian can be set up either with $$+\lambda$$ or $$-\lambda$$, depending on how you solve the budget constraint.
Actually, for the solution it does not matter if $$\lambda$$ has negative or positive sign in the equation. You can clearly see it from the formula if you expand the second term: $$\mathcal{L}(x,y, \lambda) = U(x,y) + \lambda(m -p_{x}x - p_{y}{y})= U(x,y) +\lambda m - \lambda p_{x}x - \lambda p_{y}{y}$$
$$\mathcal{L}(x,y, \lambda) = U(x,y) - \lambda(p_{x}x + p_{y}{y}-m)= U(x,y) - \lambda p_{x}x - \lambda p_{y}{y} +\lambda m$$